How much gold is deposited in 6 hours of electrolysis at 0.540 A?

[~angel~]

In the electrolysis of Au3+ (aq) solution, gold is deposited. How much gold is deposited in 6 hours by a constant current of 0.540 A?

I know you need to find the moles of e- by using It/F. But I'm not sure where to go from there.

Any help would be great.

Thanks.

 

[dextercioby]

Use Faraday's law,compute the mass of Au by making those multiplications/divisions and then compare to the fninal result.

The formulation's kinda vague,so you can leave the final answer in Kg...

Daniel.

 

[siddharth]

Consider the half-reaction

Au3+(aq) + 3e- -------------> Au(s)

Now 1 mol of Au3+ requires 3 Faradays (3*96500 C) to deposit 1 mol of Au

Therefore, if the amount of charge passed is 6*3600*.540 C, how many moles of Au will be deposited?

 

[~angel~]

Okie dokie, thanks.

 

[~angel~]

I just used n = It/zF (where z= amount of "excess" electrons). Then multiplied the answer by the molar mass of gold. That gave me 7.94g.